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2 years ago

Which statement describes one acid-base theory?

1 answer:

3 0

The answer is an acid is an H+ donor and a base is a H+ acceptor. This is in accordance to the Lewis definition of acids, which states that an acid is any substance that is able to release H+ ions in a solution or one that is able to accept a pair of nonbonding electrons. Another example of an acid-base theory is the Bronsted-Lowry theory of acids and bases.

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A 12-liter tank contains helium gas pressurized to 160 \rm atm.How many 3-liter balloons could the 12-L helium tank fill? Keep i

Vikentia [17]

Answer:

636 balloons

Explanation:

If we assume that helium gas follows an ideal gas behaviour, we can use the ideal gas law to solve this problem as follows:

We consider two different states, the initial given by the conditions of the problem statement and the final, when the tank reaches atmospheric pressure and it's no longer able to fill balloons: $P_{1}=160 atm\\V_{1}=12 L\\P_{2}=1 atm\\V_{2}= ?$
To find out what would be this volume 2, we use the Boyle's Law: $P_{1}V_{1}=P_{2}V_{2}\\V_{2}=\frac{P_{1}V_{1}}{P_{2}} \\V_{2}=\frac{160 atm \times 12L}{1 atm}\\V_{2}=1920 L$
Now we find the available volume to fill the balloons by substracting both, volume 2 and volume 1: $V_{b}=V_{2}-V_{1}=1920L-12L=1908 L$
Finally, we determine the quantity of ballons by dividing that available volume between the volume of each ballon: $B=\frac{1908L}{3L} =636 balloons$

8 0

2 years ago

A student ran the following reaction in the laboratory at 311 K:CH4(g) + CCl4(g) 2CH2Cl2(g)When she introduced 4.10×10-2 moles o

statuscvo [17]

Answer: The value of $K_{eq}$ is 0.044

Explanation:

We are given:

Initial moles of methane = $4.10\times 10^{-2}mol=0.0410moles$

Initial moles of carbon tetrachloride = $6.51\times 10^{-2}mol=0.0651moles$

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

$\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}$

So, concentration of methane = $\frac{0.0410}{1.00}=0.0410M$

Concentration of carbon tetrachloride = $\frac{0.0651}{1.00}=0.0651M$

The given chemical equation follows:

$CH_4(g)+CCl_4(g)\rightleftharpoons 2CH_2Cl_2(g)$

Initial: 0.0410 0.0651

At eqllm: 0.0410-x 0.0651-x 2x

We are given:

Equilibrium concentration of carbon tetrachloride = $6.02\times 10^{-2}M=0.0602M$

Evaluating the value of 'x', we get:

$\Rightarrow (0.0651-x)=0.0602\\\\\Rightarrow x=0.0049M$

Now, equilibrium concentration of methane = $0.0410-x=[0.0410-0.0049]=0.0361M$

Equilibrium concentration of $CH_2Cl_2=2x=[2\times 0.0049]=0.0098M$

The expression of $K_{eq}$ for the above reaction follows:

$K_{eq}=\frac{[CH_2Cl_2]^2}{[CH_4]\times [CCl_4]}$

Putting values in above expression, we get:

$K_{eq}=\frac{(0.0098)^2}{0.0361\times 0.0603}\\\\K_{eq}=0.044$

Hence, the value of $K_{eq}$ is 0.044

8 0

3 years ago

If 3 newton's of force is pulling

Troyanec [42]

A. There is no movement
Hope this helps

5 0

3 years ago

Explain why the pbcl2 dissolved when water was added

lara [203]

PbCl2 would not dissolve because it is insoluble based on the solubility rules for substances that will dissolve in water. This compound would instead form a solid precipitate at the bottom of the container.

4 0

3 years ago

What mass in grams would 5.7L of hydrogen gas occupy at STP?

tekilochka [14]

Answer: The correct answer is: " 0.54 g " .

__________________________________________

Explanation:

Note that "hydrogen gas" is:

H₂ (g) ; that is: a "diatomic element" (diatomic gas) ;

_________________________________________

The molecular weight of "H" is: 1.00794 g ;

(From the Periodic Table of Elements).

So, the molecular weight of: H₂ (g) is:

" 1.00794 g * 2 = 2.01588 g ; {use calculator) ;

_________________________________________

Note the conversion for a gas at STP:

______

1 mol of a gas = 22.4 L gas;

___

i.e. " 1 mol / 22.4 L " ;

____

So: " 5.7 L H₂ (g) $* \frac{1 mol H_{2} }{22.4 L} *\frac{2.01588 g}{mol} =? ;$

The "L" ("literes" cancel out to "1" ; since "L/L = 1 ;

The "mol" (moles) cancel out to "1" ; since "mol/mol = 1 ;

____

and we are left with:

____

[5.7 * 2.104588 g ] / 22.4 = ? g ;

______________________

→ [ 11.9961516 g ] / 22.4 =

0.53554248214 g ;l

_____________________________

We round this value to: " 0.54 g " ;

→ since "5.7 L " has 2 (two) significant figures;

22.4 is an exact number conversion;

and "5.7 L" has fewer significant figures than:

" 2.104588 " ; or: " 1.00794 " .

→ as such: We round to "2 (two) significant figures."

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Hope this is helpful. Wishing you the best in your academic endeavors!

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8 0

3 years ago